From off the top of my head, without any revision.....

IP3, or "Third Order Intercept Point" is an indication
of how good a mixer is, but it is not a physical point!

If you were to plot the wanted output of a mixer stage against
the input signal (ignoring the local oscillator input), you would
get a graph that is a nearly-straight line from the origin
which then starts to flatten off.

At the point of the line where it starts to curve over to
flatness, and therefore starts to be non-linear, other
mixer products, mainly those based upon the third
harmonic of the input signals start to appear in the
output. if you plot these other products on your graph in
addition to the wanted output signal, they grow at a rate (the slope)
which is 3 times greater than was the initial straight line
of the wanted output.

If you take the original straight line of the wanted output, and
extrapolate it so that it meets the other line growing at 3 times
the slope, you get what is known as the "Third Order Intercept
Point". The reason that this is a theoretical point is because the
wanted output has long since flattened off!

The better a mixer is, the higher is IP3 for the outputs of the mixer.

IP3 will be given in terms of the power of the wanted output signal,
say, 50 dBm - other respondents have informed you that this is
50dB (or 10^5) times greater than 1mW, or 100W (Perhaps not
a good figure for an example - a mixer with an output of those
levels could be a PA stage!). In this case dBm gives us the power relative
to the mW.

If we now go back to the flattening off of the curve, at some point,
the curve will be 1dB less than what it would have been had the curve
not been a curve but had carried on as a straight line. This point is known
as the "1dB Compression Point" - In this case we use dB and not dBm because
we are talking relative to some other point on the line.

There is a mathematical derivation (which I don't know off-hand) which
shows that the 1dB Compression Point is 10.4dB below IP3.

So, I hope that I have gone some way to explaining (or increasing your
confusion) on the points that you raised!


"jason" wrote in message
ups.com...
May I know what actually the unit of dbm and db is different from one
another?
If they are different how can we minus the gain in unit of db from a
IP3 in unit of dbm?
Kindly enlighthen
Thank you all

"Airy R.Bean" wrote -

The better a mixer is, the higher is IP3 for the outputs of the mixer.

==========================

From a circuit operational point of view, could you please summarise in what
way a high IP3 makes a better mixer?

Am I correct in assuming the device need not be a mixer? Could it be an
amplifier? In which case some of the better or worse parameters would just
become meaningless.
----
Reg.

Hi Reg -

It looks like I saw your post before Airy, so I'll respond.
(Note: I sometimes have difficulty telling when some of the posters here
are really looking for answers, and when they are just trying to bait each
other. I am going to start by assuming that you are in the former
category.)

More below...


"Reg Edwards" wrote in message
...

"Airy R.Bean" wrote -

The better a mixer is, the higher is IP3 for the outputs of the mixer.

==========================

From a circuit operational point of view, could you please
summarise in what way a high IP3 makes a better mixer?

A higher IP3 simply means that the hypothetical mixer can handle
bigger signals before it produces a given level of 3rd order IM
distortion products.

Whether or not that makes it a "better" mixer would depend on
many other specifications, such as noise figure, loss/gain, bandwidth,
and even size and power consumption.

Am I correct in assuming the device need not be a mixer?
Could it be an amplifier?

If you're asking whether other components such as amplifiers
can have IP3 specifications, the answer is definitely yes.

In which case some of the better or worse parameters would just
become meaningless.

I can't figure out what you're trying to say in this last sentence.
But correct and relevant specifications are never "meaningless."
(At least I can't think of any such pathological examples.)

----
Reg.

Hello to all that Helps
You are all very kind and genius.
It is first time I asked question in newsgroup and I received so many
kind answers.
I am so happy. Thanks a lot

So from what you have all explained, can I bold enough to say that I
can add any value in db and any value in dbm together without
converting to one db or dbm unit because they are in the ratio form and
having virtually the same 10log (P1/P2) formula and nothing else more??
Please verify

Thank you

rgds and thanks
Jason

On 4 Feb 2005 20:02:05 -0800, "jason" wrote:

Hello to all that Helps
You are all very kind and genius.
It is first time I asked question in newsgroup and I received so many
kind answers.
I am so happy. Thanks a lot

So from what you have all explained, can I bold enough to say that I
can add any value in db and any value in dbm together without
converting to one db or dbm unit because they are in the ratio form and
having virtually the same 10log (P1/P2) formula and nothing else more??
Please verify

Dear Jason,

I am sorry to say that I cannot agree with your conclusion.

In my opinion, in the general case, you can NOT add dB and dBm. They
have different applications and were not intended to be added
together.

A quantity expressed in dBm is intended to convey, or imply, a
specific power level in a specific load impedance. On the other hand
the dB is NOT so defined. Therefore, I cannot think of a situation in
which you could simply add them together and have a result that anyone
would be able to iterpret at face value.


Bob, W9DMK, Dahlgren, VA
Replace "nobody" with my callsign for e-mail
http://www.qsl.net/w9dmk
http://zaffora/f2o.org/W9DMK/W9dmk.html

Dear Bob,

I understand what you mean. But in the RF lecture, gain is given in db
unit, while IP3 is in dbm unit, then in order to find overall IP3 for a
cascaded system, the gain and IP3 which in different db unit are add or
deduct from one another without effort to change the unit. WHy is it
so?
Anyone can help?
Thank you

rgds and thanks
Jason

Dear Bob,

I understand what you mean. But in the RF lecture, gain is given in db
unit, while IP3 is in dbm unit, then in order to find overall IP3 for a
cascaded system, the gain and IP3 which in different db unit are add or
deduct from one another without effort to change the unit. WHy is it
so?
Anyone can help?
By the way, how can I get notification from this newsgroup whenever
there is new contribution? No email notify me on this. How to do the
correct setting?
Thank you

rgds and thanks
Jason

On 4 Feb 2005 21:04:19 -0800, "Jason" wrote:

Dear Bob,

I understand what you mean. But in the RF lecture, gain is given in db
unit, while IP3 is in dbm unit, then in order to find overall IP3 for a
cascaded system, the gain and IP3 which in different db unit are add or
deduct from one another without effort to change the unit. WHy is it
so?
Anyone can help?
By the way, how can I get notification from this newsgroup whenever
there is new contribution? No email notify me on this. How to do the
correct setting?


Assume that your system has IP3 = X dBm measured at point Y in the
system.

Assume further that there are amplifier stages A and B following point
Y, and that those amplifiers contribute a gain of A dB and B dB.

At the output of B the power level, in dBm, will be X + A + B.

In this way you avoid the confusion by clearly stating that the power
level at the output is X + A + B. In that context, the values can be
added, because you have made it clear that it is a power level in dBm
- not a gain in dB.

Regarding the auto-notification. I do not know of any such system.
Perhaps that is an opportunity for someone to develop a valuable
product.


Bob, W9DMK, Dahlgren, VA
Replace "nobody" with my callsign for e-mail
http://www.qsl.net/w9dmk
http://zaffora/f2o.org/W9DMK/W9dmk.html

You could start off with a signal measured in dBm, perhaps
the output of a transmitter.

Thereafter you could add and subtract anything in dB (but NOT dBm).

Here's an example (my figures are made-up and not intended to
be realistic). let us calculate the E-M-E signal strength of our
transmission bounced off the moon.

TX output = 1000W = 60dBm.

Antenna gain (Assume a big dish) = 50dB

Effected Radiated Power (ERP) = 110dBm

Path loss to Moon = 80dB

Path loss due to bouncing off Green Cheese = 30dB

Path loss back from Moon = 80 dB again

Total path loss = 190dB

Antenna Gain = 50dB

Received Signal Strength = 110 -190 + 50 = -30dBm = 1 uWatt.

So, we started off with dBm, then added or subtracted dB (which
gave us dBm again, but we only added or subtracted dB)

"jason" wrote in message
ups.com...

So from what you have all explained, can I bold enough to say that I
can add any value in db and any value in dbm together without
converting to one db or dbm unit because they are in the ratio form and
having virtually the same 10log (P1/P2) formula and nothing else more??

Hello All the Kind and Clever People

I think I got what you all explained for me. I will reread them
carefully before asking more in order to save your precious time.
I am thankful to you all.
Thank you so much for people who wrote above with great efforts

Jason